(Solution Manual) Applied Linear Algebra Instructor’s Solutions Manual

Olver,Shakiban, 1

Math & Statistics

Solution Manual

1st Edition

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Solution Manual for Applied Linear Algebra Instructor’s Solutions Manual, The downloadable file consists of 886 pages, and the total size of the file is 11.39 MB. Purchase now only for highly discounted price.

Author(s): Olver, Shakiban

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1 Applied Linear Algebra Instructor’s Solutions Manual by Peter J. Olver and Chehrzad Shakiban Table of Contents Chapter Page 1. Linear Algebraic Systems ................................ ............................ 1 2. Vector Spaces and Bases ................................ ............................. 46 3. Inner Products and Norms ................................ .......................... 78 4. Minimization and Least Squares Approximation .................. 114 5. Orthogonality ................................ ................................ .............. 131 6. Equilibrium ................................ ................................ ............... 174 7. Linearity ................................ ................................ ................... 193 8. Eigenvalues ................................ ................................ ................. 226 9. Linear Dynamical Systems ................................ ...................... 262 10. Iteration of Linear Systems ................................ ....................... 306 11. Boundary Value Problems in One Dimension .......................... 346 1 3 8 21 7 − / — − − — − − − — − 1 C A 3 3 3 3 2 2 2 2 2 2 3 3 3 4 Solutions — Chapter 1 1.1.1. (a) Reduce the system to x − y = 7, 3 y = −4; then use Back Substitution to solve for x = 17 , y = − 4 . (b) Reduce the system to 6 u + v = 5, − 5 v = 5 ; then use Back Substitution to solve for 2 2 u = 1, v = −1. (c) Reduce the system to p + q − r = 0, −3 q + 5 r = 3, − r = 6; then solve for p = 5, q = −11, r = −6. 3 (d) Reduce the system to 2 u − v + 2 w = 2, − 2 v + 4 w = 2, − w = 0; then solve for u = 1 , v = − 4 , w = 0. (e) Reduce the system to 5 x1 + 3 x2 − x3 = 9, 1 x2 − 2 x3 = 2 , 2 x3 = −2; then solve for x1 = 4, x2 = −4, x3 = −1. 5 5 5 (f ) Reduce the system to x + z − 2 w = − 3, − y + 3 w = 1, − 4 z − 16 w = − 4, 6 w = 6; then solve for x = 2, y = 2, z = −3, w = 1. (g ) Reduce the system to 3 x1 + x2 = 1, 8 x2 + x3 = 2 21 x3 + x4 = 3 55 x4 = 5 ; then solve for x1 = 3 , x2 = 2 , x3 = 2 , x4 = 3 . 11 11 11 11 1.1.2. Plugging in the given values of x, y and z gives a+2 b c = 3, a 2 c = 1, 1+2 b+c = 2. Solving this system yields a = 4, b = 0, and c = 1. 1.1.3. (a) With Forward Substitution, we just start with the top equation and work down. Thus 2 x = 6 so x = 3. Plugging this into the second equation gives 12 + 3y = 3, and so y = 3. Plugging the values of x and y in the third equation yields 3 + 4( 3) z = 7, and so z = 22. (b) We will get a diagonal system with the same solution. (c) Start with the last equation and, assuming the coefficient of the last variable is = 0, use the operation to eliminate the last variable in all the preceding equations. Then, again assuming the coefficient of the next -to-last variable is non-zero, eliminate it from all but the last two equations, and so on. (d) For the systems in Exercise 1.1.1, the method works in all cases except (c) and (f ). Solving the reduced system by Forward Substitution reproduces the same solution (as it must): (a) The system reduces to 3 x = 17 , x + 2 y = 3. 2 2 (b) The reduced system is 15 u = 15 , 3 u − 2 v = 5. (c) The method doesn’t work since r doesn’t appear in the last equation. (d) Reduce the system to 3 u = 1 , 7 u − v = 5 , 3 u − 2 w = −1. (e) Reduce the system to 2 x1 = 8 , 4 x1 + 3 x2 = 4, x1 + x2 + x3 = −1. (f ) Doesn’t work since, after the first reduction, z doesn’t occur in the next to last equation. (g ) Reduce the system to 55 x1 = 5 , x2 + 21 x3 = 3 , x3 + 8 x4 = 2 , x3 + 3 x4 = 1. 21 7 8 4 3 3 1.2.1. (a) 3 × 4, (b) 7, (c) 6, (d) ( −2 0 1 2 ), (e) 0 2 . , , ♥ 0 B@ 1 −6 2 4 5 6 B C ! B 0 B 2 C B C @ A x4 1 — − − − − 0 B 1 C B C B 1 2 y 3 (d) A = @ −1 3 3 A, x = B@ v CA, b = B@ −2 CA; (g ) A = B C, x = , b = . 5 5 5 5 5 0 0 0 1 A @ 0 0 0 0 A @ − 0 0 1.2.2. (a) 0 1 2 3 1 @ 7 8 9 A 0 1 1 , (b) 1 1 2 3 , (c) 1 4 5 7 4 , (d) ( 1 2 3 4 ), 3 (e) B@ 2 CA, (f ) ( 1 ). 1.2.3. x = − 1 , y = 4 , z = − 1 , w = 2 . 3 3 3 3 1.2.4. (a) A = 1 −1 ! , x = x ! , b = 7 ! ; (b) A = 6 1 ! , x = u ! , b = 5 ! ; 3 2 v 1 1 −1 1 5 0 p 1 0 0 1 (c) A = B@ 2 −1 3 AC, x = B@ q CA, b = B@ 3 CA; 0 −1 −1 0 r 6 B 2 1 2 1 C 0 u 1 0 3 1 0 4 −3 0 w 7 B 5 3 −1 1 C 0 x1 1 0 9 1 (e) A = @ 3 2 −1 A, x = B@ x2 CA, b = B@ 5 CA; 1 1 2 x3 0 1 0 1 −2 1 0 x 1 1 −3 1 (f ) A = B 2 −1 2 −1 C, x = B y C, b = B C; @ 1 3 2 −1 A 0 3 1 0 0 1 0 x1 B@ z CA 1 B@ 0 1 1 2 CA 1 3 1 0 x 1 1.2.5. 0 1 3 1 0 0 1 3 B@ x3 CA B@ 1 CA (a) x − y = −1, 2 x + 3 y = −3. The solution is x = − 6 , y = − 1 . (b) u + w = −1, u + v = −1, v + w = 2. The solution is u = −2, v = 1, w = 1. (c) 3 x1 − x3 = 1, −2 x1 − x2 = 0, x1 + x2 − 3 x3 = 1. The solution is x1 = 1 , x2 = − 2 , x3 = − 2 . (d) x + y z w = 0, x + z + 2 w = 4, x y + z = 1, 2 y z + w = 5. The solution is x = 2, y = 1, z = 0, w = 3. 1.2.6. 1 0 0 0 0 1 0 0 0 B 0 0 0 0 1 C (a) I = , O = @ 0 0 1 0 0 0 (b) I + O = I , I O = O I = O. No, it does not. 1.2.7. (a) unde fined, (b) unde fined, (c) 3 6 0 ! , (d) unde fined, (e) unde fined, 0 1 −1 4 2 1 3 3 0 −6 −4 2 w 1 . − C 1 A − C 2 3 4 5 6 7 8 9 0 0 0 0 0 0 3 7 8 8 @ A 1 11 9 B C 9 2 14 B C (f ) @ 3 −12 −12 A, (g ) undefined, (h) −8 6 −17 , (i) undefined. 12 −3 28 4 0 B 1 C @ A B C B − C 0 1 @ A B C × @ A 0 a 0 1 ! × × × × − − z w az bw bz bw 0 1 1 0 0 0 1 0 0 0 1.2.8. Only the third pair commute. 1.2.9. 1, 6, 11, 16. 1.2.10. (a) 1 0 0 0 0 0 , (b) 0 0 −1 2 0 0 0 0 2 0 0 . 0 0 3 0 0 0 0 −3 1.2.11. (a) True, (b) true. ♥ 1.2.12. (a) Let A = x y ! . Then AD = ax by ! = ax ay ! = DA, so if a /= b these are equal if and only if y = z = 0. (b) Every 2 × 2 matrix commutes with a 0 ! = a I . x 0 0 (c) Only 3 3 diagonal matrices. (d) Any matrix of the form A = 0 y z . (e) Let 0 u v D = diag (d1, . . . , dn). The (i, j) entry of AD is aij dj. The (i, j) entry of DA is di aij. If di /= dj, this requires aij = 0, and hence, if all the di’s are different, then A is diagonal. 1.2.13. We need A of size m × n and B of size n × m for both products to be defined. Further, AB has size m × m while BA has size n × n, so the sizes agree if and only if m = n. 1.2.14. B = x y where x, y are arbitrary. 0 x 1.2.15. (a) (A + B)2 = (A + B)(A + B) = AA + AB + BA + BB = A2 + 2AB + B2, since AB = BA. (b) An example: A = 1 2 ! , B = 0 0 ! . 1.2.16. If AB is defined and A is m n matrix, then B is n p matrix and AB is m p matrix; on the other hand if BA is defined we must have p = m and BA is n n matrix. Now, since AB = BA, we must have p = m = n. 1.2.17. A On×p = Om×p, Ol×m A = Ol×n. 1.2.18. The (i, j) entry of the matrix equation c A = O is caij = 0. If any aij the only possible way that c =/ 0 is if all aij = 0 and hence A = O. 1.2.19. False: for example, 1 0 ! 0 0 ! = 0 0 ! . 0 then c = 0, so 1.2.20. False — unless they commute: AB = BA. 1.2.21. Let v be the column vector with 1 in its jth position and all other entries 0. Then A v is the same as the jth column of A. Thus, the hypothesis implies all columns of A are 0 and hence A = O. 1.2.22. (a) A must be a square matrix. (b) By associativity, AA2 = AAA = A2 A = A3. (c) The na¨ıve answer is n 1. A more sophisticated answer is to note that you can com - pute A2 = AA, A4 = A2 A2, A8 = A4 A4, and, by induction, A2r with only r matrix multiplications. More generally, if the binary expansion of n has r + 1 digits, with s nonzero digits, then we need r + s 1 multiplications. For example, A13 = A8A4A since 13 is 1101 in binary, for a total of 5 multiplications: 3 to compute A2, A4 and A8, and 2 more to mul- tiply them together to obtain A13. 5 ! ! ! ! ! 1 1 p n 0 X n p 0 X 0 0 3 4 3 3 1 2 −1 1 — 3 1 3 0 − 1 — 3 0 0 1 , (ii) 1 0 1 1.2.23. A = 0 1 . 0 0 ♦ 1.2.24. (a) If the ith row of A has all zero entries, then the (i, j) entry of AB is ai1b1j + · · · + ainbnj = 0 b1j + · · · + 0 bnj = 0, which holds for all j, so the ith row of AB will have all 0’s. (b) If A = 1 1 , B = 1 2 , then BA = 1 1 . 1.2.25. The same solution X = −1 1 in both cases. 3 −2 1.2.26. (a) 4 5 ! , (b) 5 −1 ! . They are not the same. 1 2 −2 1 1.2.27. (a) X = O. (b) Yes, for instance, A = 1 2 ! , B = 3 2 ! , X = 1 0 ! . 0 1 −2 −1 1 1 1.2.28. A = (1/c) I when c 0. If c = 0 there is no solution. 1.2.29. (a) The ith entry of A z is 1 ai1 +1 ai2 + · · · +1 ain = ai1 + · · · +ain, which is the ith row sum. (b) Each row of W has n − 1 entries equal to 1 and one entry equal to 1 − n and so its row n n 1 1 − n sums are (n − 1) n + n = 0. Therefore, by part (a), W z = 0. Consequently, the row sums of B = AW are the entries of B z = AW z = A 0 = 0, and the result follows. 0 1 1 0 1 2 −1 1 0 1 1 0 2 1 (c) z = B@ 1 CA, and so A z = B@ 2 1 3 CA B@ 1 CA = B@ 6 AC, whil e B = AW = 1 0 −4 5 −1 1 0 2 1 3 CA B@ 3 − 3 3 CA = B@ 0 1 −1 CA, and so B z = B@ 0 CA. — 4 5 −1 1 1 3 3 3 4 −5 1 0 ♦ 1.2.30. Assume A has size m × n, B has size n × p and C has size p × q. The (k, j) entry of BC is X bklclj, so the (i, j) entry of A (BC) is X aik @ bklclj 1 X X p A = aikbklclj. On the other hand, the (i, l) entry of AB is X aikbkl, so the (i, j) entry of (AB) C is lX = 1 n @ k = 1 aikbkl 1 A clj = kX = 1 lX = 1 i= 1 aikbklclj. The two results agree, and so A (BC) = (AB) C. Remark : A more sophisticated, simpler proof can be found in Exercise 7.1.44. 1.2.31. (a) We need AB and BA to have the same size, and so this follows from Exercise 1.2.13. (b) AB − BA = O if and only if AB = BA. −1 2 ! 0 0 ! B0 0 1 1 1 C 6 1 0 0 @ −1 1 0 A (d) (i) [ c A + d B, C ] = (c A + d B)C − C(c A + d B) = c(AC − C A) + d(BC − C B) = c [ A, B ] + d [ B, C ], [ A, c B + d C ] = A(c B + d C) − (c B + d C)A 1 2 1 3 − 2 ♦ p n p ♥ 0 1 3 5 3 l= 1 k = 1 l= 1 k = 1 l= 1 (c) (i) , (iii) ; 2 4 6 = c(AB − BA) + d(AC − C A) = c [ A, B ] + d [ A, C ]. (ii) [ A, B ] = AB − BA = − (BA − AB) = − [ B, A ]. 7 h i X n X n n n n n n — − @ A × 3 −1 4 0 −20 −5 5 −18 −1 ! B 1 C ! −1 2 (iii) h [ A, B ], C i = (AB − BA) C − C (AB − BA) = ABC − BAC − C AB + C BA, h [ C, A ], B i = (C A − AC) B − B (C A − AB) = C AB − AC B − BC A + BAC, [ B, C ], A = (BC − C B) A − A (BC − C B) = BC A − C BA − ABC + AC B. Summing the three expressions produces O. ♦ 1.2.32. (a) (i) 4, (ii) 0, (b) tr(A + B) = n (aii + bii) = i= 1 iX = 1 n aii + i= 1 bii = tr A + tr B. (c) The diagonal entries of AB are X aij bji, so tr(AB) = X X aij bji; the diagonal j = 1 i= 1 j = 1 entries of BA are X bji aij, so tr(BA) = X X bji aij. Thes e double summations are clearly equal. (d) tr C = tr(AB BA) = tr AB tr BA = 0 by part (a). (e) Yes, by the same proof. ♦ 1.2.33. If b = A x, then bi = ai1x1 + ai2x2 + · · · + ainxn for each i. On the other hand, cj = (a1j , a2j , . . . , anj)T , and so the ith entry of the right hand side of (1.13) is x1ai1 + x2ai2 + · · · + xnain, which agrees with the expression for bi. 1.2.34. (a) This follows by direct computation. (b) (i) −2 1 ! 1 −2 ! = −2 ! ( 1 −2 ) + 1 ! ( 1 0 ) = −2 4 ! + 1 0 ! = −1 4 ! . 3 2 1 0 3 2 3 −6 2 0 5 −6 (ii) 1 −2 0 0 −3 −1 2 2 5 −3 0 = 1 −1 1 ( 2 5 ) + −3 −2 ! ( −3 0 ) + 0 ! ( 1 −1 ) = 2 5 ! + 6 0 ! + 0 0 ! = 8 5 ! . (iii) −6 −15 3 0 2 −2 −1 −17 B0 3 −1 1 1 C 0 B 2 3 0 1 0 3 1 0 −1 1 0 1 1 @ −1 2 1 A @ 3 −1 4 CA = B@ −1 AC( 2 3 0 ) + B@ 2 CA( 3 −1 4 ) + B@ 1 CA( 0 4 1 ) 1 1 −5 0 4 1 B0 6 9 0 1 C 1 B0 −3 1 −4 1 C 1 0 B 0 4 1 1 C −5 0 B 3 14 −3 1 C = @ −2 −3 0 A + @ 6 −2 8 A + @ 0 4 1 A = @ 4 −1 9 A. (c) If we set B = x, where x is an n 1 matrix, then we obtain (1.14). n (d) The (i, j) entry of AB is kX = 1 aikbkj. On the other hand, the (i, j) entry of ck rk equals the product of the ith entry of ck, namely aik, with the jth entry of rk, namely bkj. Summing these entries, aikbkj, over k yields the usual matrix product formula. ♥ 1.2.35. (a) p(A) = A3 − 3A + i= 1 i= 1 j = 1 2 3 0 ♥ 8 — − — − ! — − − ! −1 0 2 I , q(A) = 2A2 + I . (b) p(A) = 2 8 , q(A) = 4 6 0 −1 ! . (c) p(A)q(A) = (A3 3A + 2 I )(2A2 + I ) = 2A5 5A3 + 4A2 3A + 2 I , while p(x)q(x) = 2 x5 5 x3 + 4 x2 3 x + 2. (d) True, since powers of A mutually commute. For the particular matrix from (b), p(A) q(A) = q(A) p(A) = 2 8 . −4 −6 9 − — 1 C C @ CA — 1 C C @ CA − − 0 13 − CA CA B@ − 4 3 0 1 1 7 0 3 0 1 1 0 3 5 2 1 ˛ 1 8 3 −→ 3 −5 3 ˛ 1 26 3 ! B0 B0 ˛ ˛ ♥ 1.2.36. (a) Check that S2 = A by direct computation. Another example: S = generally, 2 times any of the matrices in part (c). (b) S2 is only defined if S is square. 2 0 . Or, more 0 2 (c) Any of the matrices ± 1 0 ! , a b ! , where a is arbitrary and bc = 1 − a2. 0 !± 1 c − a (d) Yes: for example 0 1 . 1 0 1 1 1 3 0 1 ♥ 1.2.37. (a) M has size (i+j)×(k+l). (b) M = B 1 1 3 . (c) Since matrix addition is −2 2 0 1 1 −1 done entry -wise, adding the entries of each block is the same as adding the blocks. (d) X has size k × m, Y has size k × n, Z has size l × m, and W has size l t× h n. Then AX + BZ will have size i × m. Its (p, q) entry is obtained by multiplying the p row of M times the qth column of P , which is ap1x1q + · · · + apixiq + bp1z1q + · · · + bplzlq and equals the sum of the (p, q) entries of AX and BZ. A similar argument works for the remaining three blocks. (e) For example, if X = (1), Y = ( 2 0 ), Z = 0 ! , W = 0 −1 ! , then 0 1 2 0 1 1 1 0 0 1 1 4 7 0 P = B@ 0 0 −1 CA, and so M P = B 4 5 −1 . The individual block products are 1 1 0 0 ! = 1 ! (1) + 1 −1 ! 0 ! , −2 −4 −2 0 1 −1 1 −1 ! = 1 ! ( 2 0 ) + 1 −1 ! 0 −1 ! , 0 4 1 0 1 1 0 1 3 1 C 0 ! 0 5 −1 1 0 1 1 0 1 3 1 0 −1 ! −2 AC = B@ −2 AC (1) + B@ 2 0 A 1 , @B −4 −2 AC = B@ −2 CA ( 2 0 ) + B@ 2 0 CA 1 0 . 1.3.1. (a) 1 7 ˛ 4 ! 2R 1+R 2 1 7 ˛ 4 ! −2 −9 ˛ −→ 0 5 10 . Back Substitution yields x2 = 2, x1 = −10. ˛ ! − R1+R 2 ˛ ! 0 1 −2 1 ˛ 0 1 4R 1+R3 0 1 −2 1 ˛ 0 1 3 R2+R3 0 1 −2 1 ˛ 0 1 (c) 0 2 −8 −4 5 9 8 −→ ˛ −9 0 2 −8 0 −3 13 8 −→ ˛ −9 0 2 8 0 0 1 8 CA. 0 1 4 −2 ˛ 1 1 2R 1+R2 0 1 4 −2 ˛ 1 1 −3R 1+R3 0 1 4 −2 ˛ 1 1 0 1 1 2 2 2 B@ 1 −1 1 −1 1 −1 (b) . Back Substitution yields w = 2, z = 3. Back Substitution yields z = 3, y = 16, x = 29. @B B 3 10 4 B@ 51 AC 7 R2+R3 0 1 4 −2 ˛ 1 1 (d) @ −2 0 −3 ˛ −7 CA −→ B@ 0 8 −7 ˛ −5 CA −→ 0 8 −7 ˛ −5 CA 3 −2 2 ˛ −1 ˛ 3 −2 2 ˛ −1 0 −14 8 ˛ −4 −→ 0 8 −7 0 0 − 17 −5 . Back Substitution yields r = 3, q = 2, p = −1. ˛ − 4 4 11 CA B@ CA 3 — − − − − − − 2 ! ! 10 8 3 3 −6 1 0 −6 3 1 0 0 −5 5 0 0 −5 5 ! −→ 3 3 2 3 6 3 3 3 3 6 2 3 3 3 3 6 3 2 @ −1 4 −3 A −→ @ 0 3 — 3 A. B −1 0 −1 2 C −→ B 0 3 −4 2 C −→ B 0 3 −4 2 C −→ B 0 3 −4 2 C. B @ ˛ 2 − − ! 1 0 −2 0 ˛ −1 1 0 1 0 −2 0 ˛ −1 1 (e) 0 1 0 −1 ˛ 2 C reduces to B 0 1 0 −1 ˛ C. 0 −3 2 0 −4 0 0 7 0 ˛ −5 0 0 2 −3 6 0 0 0 −5 ˛ 15 Solution: x4 = −3, x3 = − 3 , x2 = −1, x1 = −4. −1 3 −1 1 ˛ −2 1 0 −1 3 −1 1 ˛ −2 1 (f ) 1 −1 3 −1 ˛ C reduces to B 0 2 2 0 −2 C. 0 1 −1 4 4 −1 1 0 ˛ 7 CA 0 0 −2 4 0 0 0 −24 8 ˛ −48 Solution: w = 2, z = 0, y = −1, x = 1. 1.3.2. (a) 3 x + 2 y = 2, − 4 x − 3 y = −1; solution: x = 4, y = −5, (b) x + 2 y = −3, − x + 2 y + z = −6, − 2 x − 3 z = 1; solution: x = 1, y = −2, z = −1, (c) 3 x − y + 2 z = −3, −2 y − 5 z = −1, 6 x − 2 y + z = −3; solution: x = , y = 3, z = 1, (d) 2 x y = 0, x + 2 y z = 1, y + 2 z w = 1, z + 2 w = 0; solution: x = 1, y = 2, z = 2, w = 1. 1.3.3. (a) x = 17 , y = − 4 ; (b) u = 1, v = −1; (c) u = 3 , v = − 1 , w = 1 ; (d) x1 = 11 , x2 = − 10 , x3 = − 2 ; (e) p = − 2 , q = 19 , r = 5 ; (f ) a = 1 , b = 0, c = 4 , d = − 2 ; (g ) x = 1 , y = 7 , z = − 8 , w = 9 . 1.3.4. Solving 6 = a + b + c, 4 = 4 a + 2 b + c, 0 = 9 a + 3 b + c, yields a = −1, b = 1, c = 6, so y = −x2 + x + 6. 1.3.5. (a) Regular: 2 1 −→ 2 1 . 1 4 7 (b) Not regular. B0 3 −2 1 1 C 0 B 3 −2 1 1 C 3 −2 5 0 0 4 B0 1 −2 3 1 C 0 B 1 −2 3 1 C (d) Not regular: @ −2 4 −1 A −→ @ 0 0 5 A. (e) Regular: 3 −1 2 0 5 −7 B0 1 3 −3 0 1 C B0 1 3 −3 0 1 C B0 1 3 −3 0 1 C B0 1 3 −3 0 1 C @ 1.3.6. 2 3 −3 5 A @ 0 −3 3 5 A @ 0 0 −1 7 A @ 0 0 0 6 A (a) − i 1 + i 1 − i 1 1 ˛ − 3 i — i 1 + i 0 1 − 2 i 1 ˛ 1 − 2 i ; use Ba˛ ck Substitution to obtain the˛ solution y = 1, x = 1 − 2 i . 2 2 5 0 CA (c) Regular: B @ @B 0 12 A @ 4 4 2 − i 0 1 i ˛ 2 i 1 C 0 B i 0 1 − i ˛ 2 i 1 C (b) 0 2 i 1 + i −1 2 i i 2 ˛ 1 − 2 i −→ 0 2 i 1 + i 0 0 −2 − i ˛ 1 − 2 i A solution: z = i , y = − 1 − 3 i , x = 1 + i . (c) 1 − i 2 ˛ i ! −→ 1 − i 2 ˛ 3 i 1 ! ; — i 1 + i ˛ −1 0 2 i ˛ − 2 − 2 i solution: y = − 1 + 3 i , x = 1 . 2 . 2 2 @ 13 X 2 2 2 2 3 ˛ ˛ ij ij 1 + i i 2 + 2 i ˛ 0 1 0 1 + i i 2 + 2 i ˛ 0 1 (d) 1 − i 2 i 3 − 3 i i 3 − 11 i 0 CA −→ B@ 0 1 −2 + 3 i 0 0 −6 + 6 i 0 CA; solution: z = − 1 − 1 i , y = − 5 + 1 i , x = 5 + 2 i . 1.3.7. (a) 2 x = 3, − y = 4, 3 z = 1, u = 6, 8 v = − 24. (b) x = 3 , y = − 4, z = 1 , u = 6, v = − 3. (c) You only have to divide by each coefficient to find the solution. ♦ 1.3.8. 0 is the (unique) solution since A 0 = 0. ♠ 1.3.9. Back Substitution start set xn = cn/unn for i = n − 1 to 1 with increment −1 1 0 i+1 1 end set xi = ii next j @ ci − j = 1 uij xj A 1.3.10. Since a11 a12 ! b11 b12 ! = a11b11 a11b12 + a12b22 ! , 0 a22 0 b22 0 a22b22 b11 b12 ! a11 a12 ! = a11b11 a22b12 + a12b11 ! , 0 b22 0 a22 0 a22b22 the matrices commute if and only if a11b12 + a12b22 = a22b12 + a12b11, or (a11 − a22)b12 = a12(b11 − b22). 1.3.11. Clearly, any diagonal matrix is both lower and upper triangular. Conversely, A being lower triangular requires that aij = 0 for i < j; A upper triangular requires that aij = 0 for i > j. If A is both lower and upper triangular, aij = 0 for all i diagonal matrix. j, which implies A is a ♦ 1.3.12. (a) Set l = ( aij , i > j, , u = ( aij , i < j, d = ( aij , i = j, 0, i ≤ j, 0, i ≥ j, 0, i /= j. 0 0 0 1 C 0 B 3 0 0 1 C 0 B 0 1 −1 1 C (b) L = 1 0 0 A , D = @ 0 −4 0 A , U = @ 0 0 2 A . 6 2 2 @ @ u ij 6 14 − ♦ 1.3.13. −2 0 0 0 0 5 0 0 0 1 1 0 0 0 (a) By direct computation, A2 = @B 0 0 0 CA, and so A3 = O. (b) Let A have size n × n. By assumption, aij = 0 whenever i > j − 1. By induction, one proves that the (i, j) entries of Ak are all zero whenever i > j k. Indeed, to compute the (i, j) entry of Ak+1 = A Ak you multiply the ith row of A, whose first i entries are 0, 0 0 0 15 — − — − ! 2 1 C B@ CA B@ CA− B@ CA e − × − 0 − 1 −2 −1 by the jth column of Ak, whose first j k 1 entries are non -zero, and all the rest are zero, according to the induction hypothesis; therefore, if i > j k 1, every term in the sum producing this entry is 0, and the induction is complete. In particular, for k = n, every entry of Ak is zero, and so An = O. (c) The matrix A = 1 1 has A2 = O. −1 −1 1.3.14. (a) Add −2 times the second row to the first row of a 2 × n matrix. (b) Add 7 times the first row to the second row of a 2 × n matrix. (c) Add −5 times the third row to the second row of a 3 × n matrix. (d) Add 1 times the first row to the third row of a 3 × n matrix. (e) Add −3 times the fourth row to the second row of a 4 × n matrix. 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 3 0 1 0 0 0 B 1 0 1 C 1.3.15. (a) , (b) 0 0 1 0 0 0 1 1 0 B 1 0 0 1 C , (c) 0 0 1 1 0 0 0 1 , (d) 0 0 1 0 0 0 0 1 B@ 0 0 CA 1.3.16. L3 L2 L1 = @ 2 1 0 A =/ L1L2L3. B0 1 0 0 1 C B0 1 0 0 1 C 1.3.17. E3 E2 E1 = @ −2 1 0 A, E1 E2 E3 = @ −2 1 0 A. The second is easier to predict since its entries are the same as the corresponding entries of the Ei. 1.3.18. (a) Suppose that E adds c 0 times row i to row j /= i, whil e Ee adds d /= 0 times row k to row l /= k. If r1, . . . , rn are the rows, then the effect of E E is to replace (i) rj by rl + c ri + d rk for j = l; (ii) rj by rj + c ri and rl by rl + (cd) ri + d rj for j = k; (iii) rj by rj + c ri and rl by rl + d rk otherwise. On the other hand, the effect of E Ee is to replace (i) rj by rl + c ri + d rk for j = l; (ii) rj by rj + c ri + (cd) rk and rl by rl + d rk for i = l; (iii) rj by rj + c ri and rl by rl + d rk otherwise. Comparin g results, we see that E Ee = Ee E whene ver i /= l and j /= k. (b) E1E2 = E2E1, E1E3 E3E1, and E3E2 = E2E3. (c) See the answer to part (a). 1.3.19. (a) Upper triangular; (b) both special upper and special lower triangular; (c) lower triangular; (d) special lower triangular; (e) none of the above. 1.3.20. (a) aij = 0 for all i =/ j; (b) aij = 0 for all i > j; (c) aij = 0 for all i > j and aii = 1 for all i; (d) aij = 0 for all i < j; (e) aij = 0 for all i < j and aii = 1 for all i. 1.3.21. (a) Consider the product L M of two lower triangular n n matrices. The last n i entries in the ith row of L are zero, while the first j 1 entries in the jth column of M are zero. So if i < j each summand in the product of the ith row times the jth column is zero, . 2 1 ♦ 0 0 0 1 0 0 0 1 0 −2 0 1 1 2 1 1 2 1 16 A @ 2 2 6 A @ A 1 — − 1 0 0 1 B C B C B C − 0 B 1 C B@ CA — − (c) L = @ −1 1 0 A, U = @ 0 2 0 , (d) L = 1 1 0 A, U = @ 0 3 − 1 A, 3 0 − 1 3 1 2 2 3 −1 −2 1 1 0 0 −4 1 3 3 5 and so all entries above the diagonal in LM are zero. (b) The ith diagonal entry of LM is the product of the ith diagonal entry of L times the ith diagonal entry of M . (c) Special matrices have all 1’s on the diagonal, and so, by part (b), does their product. 1.3.22. (a) L = 1 0 ! , U = 1 3 ! , (b) L = 1 0 ! , U = 1 3 ! , −1 1 0 3 3 1 0 −8 1 0 0 1 C 0 B −1 1 −1 0 B 1 0 0 1 C 0 B 2 0 3 1 C 1 0 0 1 C 0 B −1 0 0 1 C B0 1 0 0 1 C 0 B 1 0 −1 1 C (e) L = @ −2 1 0 A, U = @ 0 −3 0 A, (f ) L = @ 2 1 0 , U = 0 3 4 , −1 −1 1 1 0 0 0 C 0 0 2 0 B 1 0 −1 0 1 C −3 3 0 0 0 − 13 1 0 0 0 1 0 1 0 0 B C 0 2 1 1 B C B −1 1 0 0 C −1 2 CA B@ 0 0 1 7 CA B@ −2 1 1 0 CA 2 0 1 1 −2 3 1 0 0 0 −10 1 0 0 0 B C 0 B 2 1 3 1 1 C U = 0 3 1 3 , (i) L = 2 1 0 0 , U = 0 2 − 2 2 . @ 0 0 0 1 A B@ 2 − 7 1 0 CA B@ 0 0 − 7 7 CA 1.3.23. (a) Add 3 times first row to second row. (b) Add 2 times first row to third row. (c) Add 4 times second row to third row. 1 0 1 0 0 3 0 1 3 1 0 0 7 1 1 3 (g ) L = , U = , (h) L = , 7 3 1 22 1 1 5 35 2 7 − 22 1 0 0 0 22 17 — − — − t t t B C B C t t t t t 1 2 0 B 2 1 2 2 2 3 2 2 2 2 2 t1 t2 t3 t4 = t1 1 0 0 3 1 3 2 3 3 3 4 3 1 t2 + t 1 t 2 + t2 t 1 + t 2 + t 3 1 1.3.24. (a) 1 0 0 0 2 1 0 0 18 3 4 1 0 5 6 7 1 19 , t B@B C 2 1 t 1 + t 2 1 0 0 (t3 − t1)(t3 − t2) (b) (1) Add 2 times first row to second row. (2) Add 3 times first row to third row. 20 (3) Add 5 times first row to fourth row. (4) Add 4 times second row to third row. 21 (5) Add 6 times second row to fourth row. (6) Add 7 times third row to fourth row. 22 (c) Use the order given in part (b). 23 ♦ 1.3.25. See equation (4.51) for the general case.